AmarachiApu
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Write balanced nuclear equations for each of the following processes:

(a) Alpha emission from curium-242:
[tex]\[\ce{^{242}_{96}Cm -\ \textgreater \ ^{4}_{2}He + ?}\][/tex]

(b) Beta emission from magnesium-28:
[tex]\[\ce{^{28}_{12}Mg -\ \textgreater \ ^{0}_{-1}e + ?}\][/tex]

(c) Positron emission from xenon-118:
[tex]\[\ce{^{118}_{54}Xe -\ \textgreater \ ^{0}_{1}e + ?}\][/tex]

Sagot :

GinnyAnswer
Let's write the balanced nuclear equations step-by-step for each of the given processes.

### (a) Alpha Emission from Curium-242

An alpha particle ([tex]\(_2^4He\)[/tex]) is emitted from curium-242 ([tex]\(_{96}^{242}Cm\)[/tex]). In alpha emission, the nucleus loses 2 protons and 2 neutrons. Therefore, the resulting element will have 2 fewer protons (which means it will shift 2 places back in the periodic table) and its mass number will decrease by 4.

The equation is:
[tex]\[ \mathrm{_{96}^{242}Cm} \rightarrow \mathrm{_{2}^{4}He} + \mathrm{_{Z}^{A}X} \][/tex]

Where:
- [tex]\(Z = 96 - 2 = 94\)[/tex]
- [tex]\(A = 242 - 4 = 238\)[/tex]

We identify the resulting element as plutonium ([tex]\(_{94}^{238}Pu\)[/tex]).

So, the balanced nuclear equation is:
[tex]\[ \mathrm{_{96}^{242}Cm} \rightarrow \mathrm{_{2}^{4}He} + \mathrm{_{94}^{238}Pu} \][/tex]

### (b) Beta Emission from Magnesium-28

In beta emission, a neutron in the nucleus is converted into a proton, an electron (beta particle), and an antineutrino. The number of protons increases by one, while the mass number remains the same.

The equation is:
[tex]\[ \mathrm{_{12}^{28}Mg} \rightarrow \mathrm{_{-1}^{0}e} + \mathrm{_{Z}^{A}X} \][/tex]

Where:
- [tex]\(Z = 12 + 1 = 13\)[/tex]
- [tex]\(A = 28\)[/tex]

We identify the resulting element as aluminum ([tex]\(_{13}^{28}Al\)[/tex]).

So, the balanced nuclear equation is:
[tex]\[ \mathrm{_{12}^{28}Mg} \rightarrow \mathrm{_{-1}^{0}e} + \mathrm{_{13}^{28}Al} \][/tex]

### (c) Positron Emission from Xenon-118

In positron emission, a proton in the nucleus is converted into a neutron, a positron (positive electron), and a neutrino. The number of protons decreases by one, while the mass number remains the same.

The equation is:
[tex]\[ \mathrm{_{54}^{118}Xe} \rightarrow \mathrm{_{1}^{0}e} + \mathrm{_{Z}^{A}X} \][/tex]

Where:
- [tex]\(Z = 54 - 1 = 53\)[/tex]
- [tex]\(A = 118\)[/tex]

We identify the resulting element as iodine ([tex]\(_{53}^{118}I\)[/tex]).

So, the balanced nuclear equation is:
[tex]\[ \mathrm{_{54}^{118}Xe} \rightarrow \mathrm{_{1}^{0}e} + \mathrm{_{53}^{118}I} \][/tex]
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